A car starting from rest at a stoplight accelerates to 70.0 mph in 6.5 s. Once t

Question

A car starting from rest at a stoplight accelerates to 70.0 mph in 6.5 s. Once the car reaches that speed, it is maintained for 15s until it reaches point x_2. What is the acceleration a_1 of the car in m/s^2? What is the distance x_1 covered by the car while accelerating up to 70.0 mph? What is distance between x_1 and x_2? A police officer is halfway between x_1 and x_2 and observes the car speeding past at that same point. The police officer immediately begins pursuit starting from rest and catches up to the car by point x_2. What is the acceleration a_cop of the police officer?

Explanation / Answer

A)

initial velocity, u= 0 m/s2   

Velocity of car when it reaches point x1, v = 70 mph

Converting mph to m/s2 v = 70 * 1.6 * 1000/ 3600 = 31.12 m/s2

Applying kinematics formula that is :: v= u + a * t

a= (v-u) / t

a= 31.12 / 6.5

a= 4.7863 m/s2

B)

s= u *t + 1/2 * (a*t2)

u = 0 m/s2

t= 6.5 s

s= 0 + 1/2 * (4.7863 * 6.52)

s= 101.111 m

C)

Since the speed is constant after point x1. So, distance = speed * time

distance covered = 31.12 * 15 = 466.8 m

D) Police officer starts from halfway between x1 and x2.

distance covered :: 233.4 m

time taken by police officer to catch the car is half the time taken by car to reach x2 from x1.

time =7.5 s

s= u* t + 1/2 *(a*t2)

s= 0 + 1/2 * (a * 7.5*7.5)

acop=8.29 m/s2

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